Calculation
P = U · I
I = P/U; at constant power, current falls as voltage rises.
P = U·I
Watts to Amps Calculator
I = P/U; at constant power, current falls as voltage rises.
MINTSI
02
Result
Select a target and calculate.
Technical principle schematic
Understand the inputs
- Supply voltage U — The mains or supply voltage the load runs on. Source: the installation's nameplate, or the standard socket rating (in Germany 230 V single-phase, 400 V between line conductors for three-phase).
- Current I — The operating current the wire and fuse must carry. As a result it follows from P/U; it is the target quantity for choosing cable size and fuse rating (not the other way round: the fuse is sized to this current).
- Power P — The rated or real power drawn by the load. Source: the device's nameplate (rating in W or kW).
Example
2,300 W at 230 V require 10 A (ohmic load).
Assumptions and limits
Power factor 1; motor inrush, reactive power and harmonics are excluded.
Understand Watts to Amps Calculator
This calculator finds the missing quantity among current I, voltage U and power P (plus an equivalent ohmic resistance R) from the other two. It helps convert a power rating in watts into the actual current in amps — a core step in sizing fuses and cable cross-sections.
What does this quantity describe?
A load's operating current follows from its power divided by supply voltage: I = P/U. For an ohmic load, an equivalent resistance R = U/I can additionally describe the load as a single resistor.
At a given power, required current falls as supply voltage rises — which is why long-distance power lines run at high voltage: the same transmitted power needs a smaller current at higher voltage, allowing thinner, lower-loss conductors.
Formula and variables
P = U · I
I = P / UU = P / IP = U · IR = U / I
| Symbol / input | Meaning |
|---|---|
| Supply voltage U | The mains or supply voltage the load runs on. Source: the installation's nameplate, or the standard socket rating (in Germany 230 V single-phase, 400 V between line conductors for three-phase). |
| Current I | The operating current the wire and fuse must carry. As a result it follows from P/U; it is the target quantity for choosing cable size and fuse rating (not the other way round: the fuse is sized to this current). |
| Power P | The rated or real power drawn by the load. Source: the device's nameplate (rating in W or kW). |
Choose the inputs correctly
Enter two of U, I or P; the calculator computes the third plus the equivalent ohmic resistance R. Use the actual operating voltage, not necessarily a component's printed rated voltage.
How to use the calculator
Select the target quantity and enter the two known values with units. For fuse sizing, add a safety margin to the calculated current afterward rather than sizing exactly to the raw result.
Worked example
Given P = 2,300 W and U = 230 V, substitution gives I = 2,300 W / 230 V = 10 A (for an ohmic load).
Understand the result and units
A result of 10 A means an operating current of 10 A flows at this power and voltage. Fuse and cable cross-section must be rated for at least this current — usually with a safety margin.
Current in amperes (A), voltage in volts (V), power in watts (W). The equivalent ohmic resistance is given in ohms (Ω) and is only meaningful for genuinely ohmic loads.
Why do long-distance power lines run at high voltage?
For a given power to transmit, current falls as voltage rises (I = P/U). Since power loss in a line grows with the square of current (P_loss = I²·R_line), high voltage and correspondingly low current drastically cut transmission losses. This is why high-voltage transmission lines run at several hundred thousand volts, while household circuits at 230 V need much higher currents for the same power.
Typical applications
The conversion is used to size fuses, cable cross-sections, plugs and distribution boards, and to estimate grid load from several simultaneously operating appliances.
Assumptions, limits and common mistakes
The formula assumes an ohmic load with power factor 1. For motors, power supplies and other nonlinear or inductive loads, actual current deviates from the calculated value, especially due to high inrush current at startup.
Common mistake: Do not size a fuse exactly to the calculated current: for continuous loads, the US National Electrical Code (NEC 210.19), for example, mandates a 125% margin; motors with high inrush current often need considerably more. In Europe, national standards such as IEC 60364/VDE 0100 govern cable and fuse sizing with their own factors. Also, a fuse must always protect the cable cross-section, not just the connected device.
Frequently asked questions
How large should a fuse be relative to the calculated current?
A margin is standard for continuous loads; the US National Electrical Code mandates 125%, for example (fuse rating ≈ 1.25 × calculated current). Motors with high inrush current often need considerably more.
Why does current fall at higher voltage?
Because at constant power P = U·I the product stays fixed: as U rises, I must fall correspondingly.
Does this formula apply to electric motors too?
Only approximately for continuous, ohmic-like operation. A motor's inrush current can be many times the calculated continuous current.